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At 298 K, the enthalpy change for a given reaction K is -x cal/mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature

(a) can be negative, but numerically larger than $\frac{x}{298} cal K^{-1}$
(b) can be negative, but numerically smaller than $\frac{x}{298} cal K^{-1}$
(c) cannot be negative
(d) cannot be positive

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Toolbox:
  • $\Delta G = \Delta H - T\Delta S$
Answer: can be negative, but numerically smaller than $\frac{x}{298} cal K^{-1}$
$\Delta G = \Delta H - T\Delta S$
For spontaneity, $\Delta G < 0$
As $\Delta H$ = - x kcal is negative and T = 298 K.
For $\Delta G$ to be < 0, $\Delta S$ can be negative but numerically less than $\frac{x}{298}cal K^{-1}$
answered Feb 24, 2014 by mosymeow_1
 

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